Chapter17 - 17.1. Model: For a gas, the thermal energy is...

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17.1. Model: For a gas, the thermal energy is the total kinetic energy of the moving molecules. That is, E th = K micro . Solve: The number of atoms is N M m == × 0 664 10 301 10 27 23 .0020 kg kg . . Because helium atoms have an atomic mass number A = 4, the mass of each helium atom is m × () −− 4 6 64 10 27 u 4 1.661 10 kg kg 27 . The average kinetic energy of each atom is Km v avg avg 2 kg 700 m s ==× 1 2 1 2 27 2 . = 1.63 × 10 21 J Thus the thermal energy of the gas is EK N K th micro avg J == = × × 3 01 10 1 63 10 23 21 .. = 490 J
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17.2. Model: For a gas, the thermal energy is the total kinetic energy of the moving molecules. Solve: Oxygen atoms have an atomic mass number A = 16, so the mass of each molecule is m = 32 u = 32(1.661 × 10 27 kg) = 5.32 × 10 26 kg The number of molecules in the gas is N M m == × 0 0080 26 . k g 5.32 10 kg = 1.505 × 10 23 The thermal energy is EN K N m v th avg avg 2 () 1 2 ⇒= = × × v E Nm avg th 23 1700 J 1.505 10 kg 2 2 532 10 26 . = 650 m/s
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17.3. Model: The work done on a gas is the negative of the area under the pV curve. Visualize: Please refer to Figure Ex17.3. The gas is compressing, so we expect the work to be positive. Solve: The work done on the gas is W pdV pV =− () =−− × = area under the curve 200 cm 200 kPa m Pa 40 J 33 200 10 2 0 10 65 . Assess: The area under the curve is negative because the integration direction is to the left. Thus, the environment does positive work on the gas to compress it.
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17.4. Model: The work done on a gas is the negative of the area under the pV curve. Visualize: Please refer to Figure Ex17.4. The gas is expanding, so we expect the work to be negative. Solve: The area under the pV curve is the area of the rectangle and triangle. We have 200 10 200 10 200 10 200 10 63 1 2 × () × × −− m Pa m Pa 33 = 60 J Thus, the work done on the gas is W = 60 J. Assess: The environment does negative work on the gas as it expands.
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17.5. Visualize: Please refer to Figure Ex17.5. Solve: The work done on gas in an isobaric process is Wp Vp V V =− () fi Substituting into this equation, 80 J Pa × 200 10 3 3 11 VV ⇒= × = V i 33 m cm 2 0 10 200 4 . Assess: The work done to compress a gas is positive.
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17.6. Model: Helium is an ideal gas that undergoes isobaric and isothermal processes. Solve: (a) Since the pressure ( p i = p f = p ) is constant the work done is Wp V p V V nRT V VV on gas fi i i 33 3 0.10 mol 8.31 J mol K 573 K cm cm cm J =− () = 1000 2000 2000 238 (b) For compression at a constant temperature, W nRT V V on gas 3 3 0.10 mol 8.31 J mol K 573 K m m 330 J × × = ln ln 1000 10 2000 10 6 6 (c) For the isobaric case, p nRT V == × i i Pa 238 10 5 . For the isothermal case, p i Pa 5 . and the final pressure is p nRT V f f f Pa × 476 10 5 .
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17.7. Visualize: Solve: Because Wp d V =− and this is an isochoric process, W = 0 J. The final point is on a higher isotherm than the initial point, so T f > T i . Heat energy is thus transferred into the gas ( Q > 0) and the thermal energy of the gas increases ( E th f > E th i ) as the temperature increases.
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17.8.
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This homework help was uploaded on 04/02/2008 for the course PHYS 131-133 taught by Professor All during the Spring '08 term at Cal Poly.

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Chapter17 - 17.1. Model: For a gas, the thermal energy is...

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